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我正在使用带有 Gulp 的入门主题 Roots与 MAMP PRO 合作开发一个 Wordpress 项目。我遇到了来自 gulpfile.js 的“监视”任务的问题。下面是任务的代码:

gulp.task('watch', function() {
  browserSync({
    proxy: "mywebsite:8888",
    port: 8888,
    open: "internal"
  });
  gulp.watch([path.source + 'styles/**/*'], ['styles']);
  gulp.watch([path.source + 'scripts/**/*'], ['jshint', 'scripts']);
  gulp.watch(['bower.json'], ['wiredep']);
  gulp.watch('**/*.php', function() {
    browserSync.reload();
  });
});

这是启动“gulp watch”命令后的终端输出:

gulp watch
[12:46:32] Using gulpfile ~/Dropbox/_sites/mywebsite/wp-content/themes/mywebsitetheme/gulpfile.js
[12:46:32] Starting 'watch'...
[12:46:37] Finished 'watch' after 4.26 s
[BS] Proxying: http://mywebsite:8888
[BS] Now you can access your site through the following addresses:
[BS] Local URL: http://localhost:8888
[BS] External URL: http://192.168.22.168:8888

结果 browsersync 打开,使用我的默认浏览器,url http://localhost:8888

我现在需要的是,基本上,生成的 url(由浏览器打开)是http://mywebsite:8888

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1 回答 1

2

感谢 bower 社区成员(github 问题)解决了问题我只需要删除任何配置选项并在结束标记之前添加以下代码段:

<script type='text/javascript' id="__bs_script__">//<![CDATA[
    document.write("<script async src='//HOST:PORT/browser-sync/browser-sync-client.2.1.0.js'><\/script>".replace(/HOST/g, location.hostname).replace(/PORT/g, location.port));
//]]></script>
于 2015-02-16T15:58:25.663 回答